# Fixed point at infinity

I have three questions about a fixed point at infinity.

1. How can be proved the following result? (if it is true)

If we have an entire map of degree $d\geq 2$ (i.e. such that the cardinality of the preimage of every point is $d$) such that it has a superattracting fixed point at $\infty$, then it is a polynomial of degree $d$.

2. Does every entire function with a superattractive fixed point at infinity (of degree $d$) have the form $f(z) = a_dz^d + a_{d+1}z^{d+1} + \dots$ with $a_d\neq 0$?

3. This is the version of Böttcher's Theorem that I have:

Let $f(z) = a_nz^n + a_{n+1}z^{n+1} + \dots$, where $n\geq 2$ and $a_n\neq 0$. Then there exists a local holomorphic change of coordinate $\omega = \phi(z)$ which conjugates $f$ to the $n$-th power map $\omega\mapsto \omega^n$ throughout some neighbourhood of $\phi(0) = 0$. Besides, $\phi$ is unique up to multiplication by an $(n-1)$-st root of unity.

I want to deduce that this theorem is also satisfied when the superattracting fixed point is $\infty$.

Thank you.

That $z_+=f(z)$ has a super-attracting fixed point at infinity can only be defined as that $w_+=g(w)$ with $g(w)=1/f(1/w)$ has a super-attracting fixed point of the same degree at zero. That is equivalent to the existence and non-nullity of the limits $$\lim_{w\to 0}\frac{g(w)}{w^d}\ne 0\iff \lim_{z\to\infty}\frac{f(z)}{z^d}\ne 0$$
1. if $f$ is entire then this kind of polynomial growth requires it to be a polynomial of degree $d$.
2. No. $\deg f=d$ says the exact opposite. However $g(w)=b_dw^d+b_{d+1}w^{d+1}+\dots$
3. This in turn means that there is a coordinate change for $|w|<r$ so that $\phi(g(w))=\phi(w)^d$ so that with $\psi(z)=1/\phi(1/z)$ for $|z|>R$ $$\psi(f(z))=\psi(z)^{d}$$ with the same uniqueness properties. This means nothing more than that outside a large enough disk $B(0,R)$ the terms of $f(z)$ besides $a_dz^d$ are negligible, and that in the rescaled equation $cz_+=(a_d/c^{d-1})(cz^d)$ has $d-1$ solutions for $c^{d-1}=a_d$ to find the normal form with leading coefficient $1$.
• Thank you @LutzL . I was not acquainted with infinity as a fixed point and now I understand it. Only a little doubt: at the last part, when you write the rescaled equation, do you mean just to consider the change $f(z) \to cf(z/c)$? – dudas Apr 2 '18 at 12:58
• Essentially yes. You can transform $z_+=f(z)$ by inserting $u=cz+m$ to get the transformed recursion $u_+=cf((u-m)/c)+m$ which does not change the qualitative behavior of the recursionm the polynomial nature of $f$ nor the degree and try to fix the parameters $c,m$ so that some of the transformed coefficients have some standard value, $0$ or $1$. – Lutz Lehmann Apr 2 '18 at 16:34